poj 2151 Check the difficulty of problems 概率dp

Check the difficulty of problems

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 5636 Accepted: 2474

Description

Organizing a programming contest is not an easy job. To avoid making the problems too difficult, the organizer usually expect the contest result satisfy the following two terms: 
1. All of the teams solve at least one problem. 
2. The champion (One of those teams that solve the most problems) solves at least a certain number of problems. 

Now the organizer has studied out the contest problems, and through the result of preliminary contest, the organizer can estimate the probability that a certain team can successfully solve a certain problem. 

Given the number of contest problems M, the number of teams T, and the number of problems N that the organizer expect the champion solve at least. We also assume that team i solves problem j with the probability Pij (1 <= i <= T, 1<= j <= M). Well, can you calculate the probability that all of the teams solve at least one problem, and at the same time the champion team solves at least N problems? 

Input

The input consists of several test cases. The first line of each test case contains three integers M (0 < M <= 30), T (1 < T <= 1000) and N (0 < N <= M). Each of the following T lines contains M floating-point numbers in the range of [0,1]. In these T lines, the j-th number in the i-th line is just Pij. A test case of M = T = N = 0 indicates the end of input, and should not be processed.

Output

For each test case, please output the answer in a separate line. The result should be rounded to three digits after the decimal point.

Sample Input

2 2 20.9 0.91 0.90 0 0

Sample Output

0.972

Source

POJ Monthly,鲁小石

给出m个人 t道题 至少一人做对n道题的概率

给出方程

f[i][j]表示第i个队伍做对第j题的概率。g[i][j][k]表示第i个队伍对于前j题而言做对k道的概率。

g[i][j][k] = g[i][j - 1][k - 1] * (f[i][j]) + g[i][j - 1][k] * (1 - f[i][j]);

#include <iostream>#include <string.h>#include <stdlib.h>#include <fstream>#include <stdio.h>#include <algorithm>#include <math.h>#include <map>#include <set>#include <vector>#include <queue>#include <stack>using namespace std;#define Maxn 1010#define Maxm 32double f[Maxn][Maxm];double g[Maxn][Maxm][Maxm];int m,t,n;int main(){while(cin>>m>>t>>n){if(m==0&&t==0&&n==0) break;for(int i=1;i<=t;i++){for(int j=1;j<=m;j++){cin>>f[i][j];}}memset(g,0,sizeof(g));for(int i=1;i<=t;i++){g[i][0][0]=1;for(int j=1;j<=m;j++){g[i][j][0]=g[i][j-1][0]*(1-f[i][j]);for(int k=1;k<=j;k++){g[i][j][k]=g[i][j-1][k-1]*f[i][j]+g[i][j-1][k]*(1-f[i][j]);}}}double res=1;for(int i=1;i<=t;i++){res*=(1-g[i][m][0]);}double temp=1;for(int i=1;i<=t;i++){double cnt=0;for(int j=1;j<n;j++){cnt+=g[i][m][j];}temp*=cnt;}printf("%.3lf\n",res-temp);//cout<<res-temp<<endl;}return 0;}